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If $f(x)=\int \frac{d x}{\left(x^2+2\right)}$ and $f(\sqrt{2})=0$, then $f(0)=$
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Verified Answer
The correct answer is:
$\frac{-\pi}{4 \sqrt{2}}$
$\quad f(x)=\int \frac{d x}{x^2+(\sqrt{2})^2}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+c$ Since at $x=\sqrt{2}, f(x)=0$
$$
\Rightarrow 0=\frac{1}{\sqrt{2}} \tan ^{-1}(1)+c \Rightarrow c=-\frac{\pi}{4 \sqrt{2}}
$$
Hence
$$
f(0)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{0}{\sqrt{2}}\right)+\left(-\frac{\pi}{4 \sqrt{2}}\right)=-\frac{\pi}{4 \sqrt{2}}
$$
$$
\Rightarrow 0=\frac{1}{\sqrt{2}} \tan ^{-1}(1)+c \Rightarrow c=-\frac{\pi}{4 \sqrt{2}}
$$
Hence
$$
f(0)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{0}{\sqrt{2}}\right)+\left(-\frac{\pi}{4 \sqrt{2}}\right)=-\frac{\pi}{4 \sqrt{2}}
$$
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